Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $z = \dfrac{-3t^2 - 18t}{t + 5} \div \dfrac{t^3 - 3t^2 - 54t}{t^2 - 13t + 36} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-3t^2 - 18t}{t + 5} \times \dfrac{t^2 - 13t + 36}{t^3 - 3t^2 - 54t} $ First factor out any common factors. $z = \dfrac{-3t(t + 6)}{t + 5} \times \dfrac{t^2 - 13t + 36}{t(t^2 - 3t - 54)} $ Then factor the quadratic expressions. $z = \dfrac {-3t(t + 6)} {t + 5} \times \dfrac {(t - 9)(t - 4)} {t(t - 9)(t + 6)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-3t(t + 6) \times (t - 9)(t - 4) } {(t + 5) \times t(t - 9)(t + 6) } $ $z = \dfrac {-3t(t - 9)(t - 4)(t + 6)} {t(t - 9)(t + 6)(t + 5)} $ Notice that $(t - 9)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-3t\cancel{(t - 9)}(t - 4)(t + 6)} {t\cancel{(t - 9)}(t + 6)(t + 5)} $ We are dividing by $t - 9$ , so $t - 9 \neq 0$ Therefore, $t \neq 9$ $z = \dfrac {-3t\cancel{(t - 9)}(t - 4)\cancel{(t + 6)}} {t\cancel{(t - 9)}\cancel{(t + 6)}(t + 5)} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $z = \dfrac {-3t(t - 4)} {t(t + 5)} $ $ z = \dfrac{-3(t - 4)}{t + 5}; t \neq 9; t \neq -6 $